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(C) The electronic configurations of the given metals are:$Cr: [Ar] \, 3d^5 \, 4s^1$$V: [Ar] \, 3d^3 \, 4s^2$$Mn: [Ar] \, 3d^5 \, 4s^2$$Fe: [Ar] \, 3d

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$Mn$ $(Z=25)$

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(C) The electronic configurations of the given metals are:$Cr: [Ar] \, 3d^5 \, 4s^1$$V: [Ar] \, 3d^3 \, 4s^2$$Mn: [Ar] \, 3d^5 \, 4s^2$$Fe: [Ar] \, 3d^6 \, 4s^2$To find the third ionization enthalpy,we remove two electrons first:For $Mn$,the configuration after removing two electrons $(Mn^{2+})$ is $[Ar] \, 3d^5$. This is a half-filled stable $d$-orbital configuration.Removing the third electron from this stable $3d^5$ configuration requires a very high amount of energy.Therefore,$Mn$ has the highest third ionization enthalpy among the given options.

Which of the following metals is expected to have the highest third ionization enthalpy?

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